Question

Round answers to 1 decimal place. What is the percent yield if 6.9g of hydrogen react with nitrogen to form 20.2g of ammonia

Balance the equation before you solve: H2 + N2 -> NH3

Answer 1

To calculate the percent yield, we need to first calculate the theoretical yield, which is the amount of ammonia that would be produced if the reaction were 100% efficient.

The balanced chemical equation is:

N2 + 3H2 → 2NH3

The molar mass of H2 is 2 g/mol, and the molar mass of N2 is 28 g/mol. Therefore, the number of moles of H2 and N2 in the reaction are:

• moles of H2 = 6.9 g / 2 g/mol = 3.45 mol
• moles of N2 = 20.2 g / 28 g/mol = 0.721 mol

According to the balanced equation, 3 moles of H2 react with 1 mole of N2 to produce 2 moles of NH3. Therefore, the theoretical yield of NH3 is:

• theoretical yield = 3.45 mol H2 × (2 mol NH3 / 3 mol H2) = 2.3 mol NH3

The molar mass of NH3 is 17 g/mol. Therefore, the theoretical yield of NH3 in grams is:

• theoretical yield = 2.3 mol NH3 × 17 g/mol = 39.1 g NH3

The percent yield is then calculated by dividing the actual yield (20.2 g) by the theoretical yield (39.1 g) and multiplying by 100:

• percent yield = (20.2 g / 39.1 g) × 100% ≈ 51.7%

Therefore, the percent yield of the reaction is approximately 51.7%.